From - Thu Feb 5 10:20:53 1998
From: mwdaly@pobox.com (Matthew Daly)
Newsgroups: rec.puzzles
Subject: Re: A good puzzle!
Date: Wed, 04 Feb 1998 15:29:05 GMT
Organization: Eastman Kodak Company
Lines: 84
I'll never forget the time that "GSTATH" said:
>You are given 6 cubes (A, B, C, D, E and F).
>
>You know that one of the cubes A, B and C weighs 2Kg, the other 4Kg and the
>third 6Kg, but you do NOT know which of them weighs 2Kg, 4Kg or 6Kg.
>
>You also know that the possible weights of the cubes D, E and F are 1Kg,
>3Kg, and 5Kg. Again you do not know the exact weights of the three cubes.
>
>You are also given a scale balance.
>
>The question is:
>
>Can you find out the weight of each of the 6 cubes in only 3 weighings? If
>your answer is yes then how can you achieve that?
>
>If your answer is no then which is the minimum number of weighings required?
SPOILER
As several folks (including myself) have pointed out, at least four
weighings are necessary. This post will demonstrate that four weighings
are sufficient.
First, weigh AD vs. BE, and then weigh AE vs. BF. There are nine
possible results for these two experiments; here they are with the
possible weight configurations that produce them (the code here is that
AD-0 means that the first weighing found AD heavier and the second
weighing was even):
0-0: 135642 315246 351642 531246
0-AE: 315462 513264 531462
0-BF: 135426 153624 351426
AD-0: 315624 513426 531624
AD-AE: 315642 513246 513462 513624 513642 531642
AD-BF: 135624 315426 351624 531426
BE-0: 135264 153462 351264
BE-AE: 135462 153264 315264 351462 531264
BE-BF: 135246 153246 153426 153642 351246
Splitting up the these remaining cases is left to the reader. It gets
tedious to devise separate weighing strategies for each subcase, but you
have two additional weighings, which is plenty of room to differentiate.
To tackle the biggest subcase as an example, in the AD-AE case, weigh AC
vs. EF. The worst case from that is that AC is heavier, which could be
315642, 513624, or 513642. Weighing AF vs. BD fourth will split that
into individual cases. If AC vs. EF were even, then it's either 513462
or 531642, and A vs. CF will determine which. Finally, if AC vs. EF had
EF heavier, then you'd know that it must be 513246.
-Matthew
--
Matthew Daly mwdaly@pobox.com
My opinions may have changed, but not the fact that I am right - Ashleigh Brilliant
The views expressed in this post are not necessarily those of my employer, of course.
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From - Thu Feb 5 13:42:03 1998
From: kdg977t@cnas.smsu.edu (Green Kevin D)
Newsgroups: rec.puzzles
Subject: Re: A good puzzle!
Date: 5 Feb 1998 07:57:01 GMT
Organization: Southwest Missouri State University
GSTATH (anti.spam.ch92616@ntral.ntua.gr) wrote:
: I 've been trying to solve a puzzle for some months and it seems that I am
: clueless. I think that it is an interesting and difficult one, so I am
: sending it to you asking for some help.
: ----------------------
: You are given 6 cubes (A, B, C, D, E and F).
: You know that one of the cubes A, B and C weighs 2Kg, the other 4Kg and the
: third 6Kg, but you do NOT know which of them weighs 2Kg, 4Kg or 6Kg.
: You also know that the possible weights of the cubes D, E and F are 1Kg,
: 3Kg, and 5Kg. Again you do not know the exact weights of the three cubes.
: You are also given a scale balance.
: The question is:
: Can you find out the weight of each of the 6 cubes in only 3 weighings? If
: your answer is yes then how can you achieve that?
: If your answer is no then which is the minimum number of weighings required?
SPOILER
Matthew Daly has already pointed out that 3 weighings is insufficient.
It is possible to do it in 4 weighings. I suspect that there is a
solution that specifies all four weighings ahead of time. This is
not that solution.
First weigh AD vs BE and AE vs BD.
Case 1 AD < BE and AE < BD
Weigh D vs E and E vs F
Case 2 AD < BE and AE = BD
Weigh B vs DE and B vs C
Case 3 AD < BE and AE > BD
Weigh A vs B and B vs C
Case 4 AD = BE and AE < BD
Weigh B vs DE and B vs C
Case 5 AD = BE and AE > BD
Weigh A vs DE and A vs C
Case 6 AD > BE and AE < BD
Weigh A vs B and B vs C
Case 7 AD > BE and AE = BD
Weigh A vs DE and A vs C
Case 8 AD > BE and AE > BD
Weigh D vs E and E vs F
While I'm sure that there are probably nicer solutions, this one does
show that the problem is solvable with 4 weighings.
Kev